= Solution
The <Gauss map> sends $p$ to the chosen <unit normal> $N(p)$. Differentiating $\langle N,N\rangle=1$ shows that $dN_p$ maps $T_pS$ into itself. The <shape operator> $S_p=-dN_p$ is self-adjoint because its associated <second fundamental form> is symmetric. Its <eigenvalue>[eigenvalues] $k_1,k_2$ are the <principal curvature>[principal curvatures]. A point is <umbilical point>[umbilical] when $k_1=k_2$, and the surface is <minimal surface>[minimal] when its <mean curvature> $H=(k_1+k_2)/2$ vanishes everywhere.
In an orthonormal principal basis, $dN_p$ has diagonal matrix $\operatorname{diag}(-k_1,-k_2)$. It is conformal exactly when
$$
\langle dN_pu,dN_pv\rangle=\lambda(p)\langle u,v\rangle,
\qquad \lambda(p)>0,
$$
which is equivalent to $k_1^2=k_2^2\ne0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.
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